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This is a follow-up to Probability Question: Bridge problem.

There are $n$ islands in the ocean. Each island is linked by a single bridge to each other island. The probability of each bridge collapsed independently by earthquake is $p$. Suppose there is an earthquake causing some bridges to collapse - what is the probability of the remaining bridges are able to be traversed to each and every island?

What is the correct approach to solve this problem since my original approach (underlined in the linked question) is incorrect?

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Venus
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    You don't need to have each island linked by a single bridge between each and every unique pair of islands in order to ensure no island is isolated from the others. Can you please clarify this? – barak manos Aug 24 '14 at 10:53
  • @barakmanos Each island is connected by a single bridge to the other islands – Venus Aug 24 '14 at 11:14
  • How does this differ from your earlier question? Voting to close as a duplicate. – TonyK Aug 24 '14 at 18:18
  • @TonyK The other question asks to check a specific approach (which happened to be faulty), this one asks to solve the problem. – Did Aug 25 '14 at 11:30
  • @TonyK Please don't blame because I make 2 similar OP. I am not intended to. You may want to see the comment below Did's answer. – Venus Aug 25 '14 at 15:18
  • See https://math.stackexchange.com/questions/584228/exact-probability-of-random-graph-being-connected – Henry Mar 31 '19 at 10:42

2 Answers2

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This is the (much studied) Erdős–Rényi model of random graph. Exact results for fixed values of $n$ are seldom. Let $\epsilon\gt0$. If $1-p\gt(1+\epsilon)\ln n/n$ then the graph is connected with probability converging to $1$ when $n\to\infty$. If $1-p\lt(1-\epsilon)\ln n/n$ then the graph contains isolated vertices, and thus is disconnected, with probability converging to $1$ when $n\to\infty$. This is why $\ln n/n$ is called a sharp threshold for connectedness in this model.

Did
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  • I don't get it your answer. Let's make an example. Suppose the number of island is 10 and the probability p = 1/2. Using Erdős–Rényi model, is the probability of the event $\frac{1}{2^{45}}$, where 45 comes from $\binom{10}{2}$? – Venus Aug 24 '14 at 11:13
  • Which part of "Exact results for fixed values of n are seldom" do you fail to understand? (See also @TonyK's comment to your previous question.) – Did Aug 24 '14 at 11:20
  • Could you give me an answer from the example in my previous comment (for 10 islands and p=1/2)? I'm kind of person "learning by doing" so I will immediately understand by examples – Venus Aug 24 '14 at 11:29
  • Let me repeat: it is an illusion to believe that simple exact formulas can exist for every finite value of n (except of course for n=2, 3, 4 or, say, 5). Which is the reason why I provided the only results in the field, asymptotic when n goes to infinity. (Did you by any chance check the WP page I linked to? Your comments sound as if you did not.) – Did Aug 24 '14 at 12:13
  • I did check it, but I didn't understand. Sorry for my stupidity – Venus Aug 24 '14 at 12:15
  • Stupidity $\ne$ willingness to read. – Did Aug 24 '14 at 12:16
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This post expands upon Did's post. As Did mentioned, there is no nice closed form for this probability for general $n$. However, this probability does satisfy a "nice" recurrence (nice in the eye of the beholder, I guess). Let $P(n,p)$ be the probability that $G(n,p)$ is connected, which is the value that you want. Then $P(n,p)$ satisfies the equation \begin{equation*} \sum_{k=1}^{n}{n-1 \choose k-1} P(k,p) (1-p)^{k(n-k)} = 1 \end{equation*} or rewritten \begin{equation*} P(n,p) = 1 - \sum_{k=1}^{n-1} {n-1 \choose k-1} P(k,p) (1-p)^{k(n-k)}. \end{equation*}

EDIT: I should have mentioned who first derived this formula. Gilbert (1959) derived this recurrence relation among other results in his landmark paper introducing the random graph model $G(n,p)$.

Reference: Gilbert, E. N. (1959). Random graphs. The Annals of Mathematical Statistics, 1141-1144. http://www.jstor.org/stable/2237458

D Poole
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  • Thanks! This is exactly what I am looking for :) – Venus Aug 24 '14 at 14:43
  • Venus: Eeeh? "This is exactly what I am looking for" This does not show in your question... – Did Aug 24 '14 at 16:25
  • @DPoole +1. Nice recursion relation. – Did Aug 24 '14 at 16:28
  • @Did Sorry, please don't take this personal but I was looking for "what is the probability of the remaining bridges are able to be traversed to each and every island?" I've stated it in my OP and this answer provides the formula to obtain it but your answer not :) – Venus Aug 24 '14 at 16:39
  • @DPoole Could you please share me link of websites or papers to prove formula of $P(n,p)$? Thanks :) – Venus Aug 24 '14 at 16:57
  • Venus: Do not worry about me taking this "personal" (an odd conception, I must say), and please tell me what percentage is $P(300,.9)$ then? A: 100%. B: 90%. C: 70%. D: 50%. E: 30%. F: 10%. G: 0%. The exact recursion will NOT give you the answer except maybe at a huge computational cost. Although, by the considerations in my post, the answer is clearly G. :-) You declare that "this answer provides the formula to obtain it", I much prefer the cautious, and more accurate, formulation of the OP... – Did Aug 24 '14 at 17:15
  • @Did What kind of government builds 44,850 bridges to connect 300 islands but having probability of collapsing 90%? :) – Venus Aug 25 '14 at 15:30
  • Venus: I am not sure I understand your last comment... Am I supposed to? – Did Aug 25 '14 at 17:44